Chapter 4. Clustering and classification
In this chapter we will learn how to classify data using linear discriminant analysis and k-means, one of the most known methods of clustering.
3.1 Description and overview of the data
We will use the dataset called Boston from the MASS package. The data has 506 observations from 14 variables. The dataset gives different housing values in suburbs of Boston. E.g.: per capita crime rate by town, proportion of industries per town, nitrogen oxides concentration, etc. More information about the data can be found here: https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html
library(MASS)
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14Next we can see a graphical overview of the data and the summaries of the variables.
Because we have so many variables, it would be hard to do a scatter plot for each combination of variables, an easier way to see what is the relationship between variables is making a correlation matrix.
In our matrix we can see visually which variables are more related and if they are positively or negatively related (with the colours and size of the circles), and we can also see the correlations numerically.
The highest correlation we find it between rad and tax, which means index of accessibility to radial highways and full-value property-tax rate, so obviously, the tax rate depends a lot on whether the property is easily accessible by radial highways or not. We can see how many other variables are also well correlated in the graphic.
install.packages(“corrplot”) install.packages(“tidyverse”)
library(corrplot)## corrplot 0.84 loadedlibrary(tidyverse)## -- Attaching packages ---------------------------------------------------------------------------------------- tidyverse 1.2.1 --## v ggplot2 2.2.1 v purrr 0.2.4
## v tibble 1.3.4 v dplyr 0.7.4
## v tidyr 0.7.2 v stringr 1.2.0
## v readr 1.1.1 v forcats 0.2.0## -- Conflicts ------------------------------------------------------------------------------------------- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()
## x dplyr::select() masks MASS::select()cor_matrix <- cor(Boston) %>% round(digits=2)
corrplot.mixed(cor_matrix, lower.col = "black", number.cex = .6)
We can see a summary of the variables in the following tables. For example for variable crime we can see the mean crime rate per capita is 3.6 and the max 88.9 which looks quite alarming. Or the mean concentration of nitrogen oxides is 0.55ppm.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00Finally in the following graphs we can visually see how the variables are distributed.
gather(Boston) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()
3.2 Scaling the data
Linear Discriminant analysis makes the assumption that variables are normally distributed and the variances of each variable is the same, that is why it is better to scale the data before fitting our model.
We can see next the summaries of the scaled data. Note how the values have changed, compared to the previous summary. Now the mean for each of the variables is 0. So that indicates that all the variables have the same scale now.
After scaling, we change the object to data frame format, so we can use it later as data.
boston_scaled <- scale(Boston)
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865boston_scaled <- as.data.frame(boston_scaled)3.3 Linear discriminant analysis
First we will create a categorical variable of the crime rate in the Boston dataset. For that we will use the quantiles as the break points in the categorical variable.
# First we create the quantile vector of crim:
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# Then we create the categorical variable "crime":
crime <- cut(boston_scaled$crim, breaks = bins, inlcude.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))
table(crime)## crime
## low med_low med_high high
## 126 126 126 127Then we will remove the old variable “crim” and add the new one “crime” to the dataset.
boston_scaled <- dplyr::select(boston_scaled, -crim)
boston_scaled <- data.frame(boston_scaled, crime)Later, we divide the dataset to train and test sets, so that 80% of the data belongs to the train set. This will allow us to make predictions later in this chapter.
n <- nrow(boston_scaled)
# Choose randomly 80% of the rows:
ind <- sample(n, size = n*0.8)
# create the train set;
train <- boston_scaled[ind,]
# create the test set:
test <- boston_scaled[-ind,]Now we fit the linear discriminant analysis (LDA) on the train dataset, the target variable will be crime and all the other variables as predictors.
lda.fit <- lda(crime ~., data = train)
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2555831 0.2258065 0.2630273 0.2555831
##
## Group means:
## zn indus chas nox rm
## low 0.93790425 -0.9286343 -0.195880519 -0.8804640 0.4497423
## med_low -0.09121565 -0.2392259 -0.012740041 -0.5800215 -0.1358458
## med_high -0.39744090 0.1288941 0.173380393 0.3389491 0.1466049
## high -0.48724019 1.0170891 -0.004759149 1.0582308 -0.3105911
## age dis rad tax ptratio
## low -0.8950042 0.8973709 -0.6875068 -0.7229791 -0.45078360
## med_low -0.3730231 0.3668279 -0.5527736 -0.4557369 0.01597098
## med_high 0.3649601 -0.3452487 -0.3892189 -0.3183968 -0.24004509
## high 0.8142857 -0.8394101 1.6384176 1.5142626 0.78111358
## black lstat medv
## low 0.3799121 -0.7692630 0.517175191
## med_low 0.2915903 -0.1254296 -0.008824296
## med_high 0.1013844 -0.0055812 0.202404287
## high -0.7900066 0.8027613 -0.611609837
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.09619542 0.734600214 -0.78155745
## indus -0.01721190 -0.335453677 0.53521330
## chas -0.06908586 -0.108523300 0.16343145
## nox 0.39775345 -0.630363942 -1.34768421
## rm -0.08341965 -0.088632272 -0.15449242
## age 0.29774075 -0.314026058 -0.20055612
## dis -0.07318173 -0.283142844 0.21461874
## rad 2.99582462 0.783688687 -0.13206428
## tax -0.03479838 0.190230600 0.50501699
## ptratio 0.13361484 -0.013000027 -0.11225721
## black -0.14136572 0.009807584 0.04017724
## lstat 0.21174387 -0.256173649 0.38435897
## medv 0.19160645 -0.390985505 -0.10668182
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9467 0.0400 0.0133Finally we draw the LDA biplot to visualize how the data is distributed. Each class have a different colour and the arrows represent the predictor variables, whose lenght and direction are based on the coefficients and show their impact on the model.
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col=classes, pch= classes)
lda.arrows(lda.fit, myscale = 1)
3.4 Predict LDA
First we will save the crime categories from the test set and then remove the crime variable from the test dataset.
# we save the correct classes from test data:
correct_classes <- test$crime
# remove the crime variable from the test data:
test <- dplyr::select(test, -crime)Based on the trained model, the LDA model we created calculates the probability of new observations to belong in each of the classes and it classifies it to the class with the highest probability.
We predict the classes with LDA model on the test data.
lda.pred <- predict(lda.fit, newdata = test)Finally we cross tabulate the results of our predictions using the crime categories from the test set that we saved at the beginning of this 3.4. We can observe how most of predictions are correct, only 24 of the predictions are incorrect. If we look at the probabilities, we can see how 75.4% of the predictions are correct.
table(correct = correct_classes, predicted = lda.pred$class) ## predicted
## correct low med_low med_high high
## low 15 7 1 0
## med_low 6 18 11 0
## med_high 0 3 17 0
## high 0 0 0 24table(correct = correct_classes, predicted = lda.pred$class) %>% prop.table() %>% addmargins()## predicted
## correct low med_low med_high high Sum
## low 0.147058824 0.068627451 0.009803922 0.000000000 0.225490196
## med_low 0.058823529 0.176470588 0.107843137 0.000000000 0.343137255
## med_high 0.000000000 0.029411765 0.166666667 0.000000000 0.196078431
## high 0.000000000 0.000000000 0.000000000 0.235294118 0.235294118
## Sum 0.205882353 0.274509804 0.284313725 0.235294118 1.0000000003.5 K-means clustering
First we scale the dataset Boston.
boston_scaled2 <- scale(Boston)
summary(boston_scaled2)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865# change the object to data frame
boston_scaled2 <- as.data.frame(boston_scaled2)We calculate the distances between the observations. We use the most common, euclidean distance.
dist_eu <- dist(boston_scaled2)
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970Then we run the k-means algorithm on the dataset. First we put a random number of clusters or initial cluster centers.
km <- kmeans(boston_scaled2, centers = 3)Now we will investigate what is the optimal number of clusters. For this purpose we will first set a max number of clusters to 10. Then we will use one of the most used methods for deciding the number of cluster, sum of squares.The sum of squares or total within cluster sum of squares (TWCSS), is the sum of within cluster sum of squares (WCSS). So when you plot the number of clusters and the total WCSS, the optimal number of clusters is when the total WCSS drops radically.
# avoid the kmeans to give us every time different results
set.seed(123)
# set the maximum number of clusters to 10
k_max <- 10
# calculate the total sum of squares:
twcss <- sapply(1:k_max, function(k){kmeans (boston_scaled2, k)$tot.withinss})
# see graphically the optimal number of clusters:
qplot(x = 1:k_max, y = twcss, geom = 'line')
From the above graph we can prove how the optimal number of clusters is 2. Soo we run the k means algorithm again and visualize the results. First we see a matrix with all the variables and then zoomed for variables from 1 to 5 and from 6 to 10. On the following graphics we can observe the two clusters on the data, in red and black, and how they relate for each combination of variables
km <- kmeans(boston_scaled2, centers = 2)
pairs(boston_scaled2, col=km$cluster)
pairs(boston_scaled2[1:5], col=km$cluster)
pairs(boston_scaled2[6:10], col=km$cluster)
Bonus
First we scale the Boston dataset.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00boston_scaled3 <- scale(Boston)
summary(boston_scaled3)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865Then we perform k-means.
km2 <- kmeans(boston_scaled3, center = 3)
boston_scaled3 <- as.data.frame(boston_scaled3)Then we perform LDA using the clusters as target classes.
lda.fit2 <- lda(km2$cluster ~., data=boston_scaled3)Visualize the results. We can observe how the most influential variables which separate the clusters are accessibility to radial highways (rad), nitrogen oxides concentration (nox) and proportion of residential land zoned for lots over 25,000 sq.ft (zn).
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(km2$cluster)
# plot the lda results
plot(lda.fit2, dimen = 2, col=classes, pch= classes)
lda.arrows(lda.fit, myscale = 1)
Super bonus
First we run the code below for the scaled train data that we used to fit the LDA. The code creates a matrix product, which is a projection of the data points.
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)Then we draw a plot.
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutplot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')Then we draw the same plot put assigning the colors to be the crime classes of the train set.
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)Finally, we draw the same plot but now setting the colors to be the clusters of the k-means.
{r 3d plot3} plot_ly(x = matrix_product\(LD1, y = matrix_product\)LD2, z = matrix_product\(LD3, type= 'scatter3d', mode='markers', color = km\)cluster)
In the second graphic we can identify which observations belong to which crime classes and we can observe there is some kind of clustering within classes, especially with the class high, there is a clear cluster. The rest are not as clear.
In the last graphic we should observe the graphic where each color is a cluster, and as we could have guessed, the optimal is only 2 clusters. One for high and another one for the rest of crime classes. For some reason we could not proceed to draw this graphic.